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Math Help - Vector

  1. #1
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    Vector

    A straight line is given as 2(x+4) = -2(y-5) = -3(z-6),

    (a)determine the vector equation of the straight line. The answer I found is r = -4i +5j +6k + t(3i -3j +2k), is this correct?
    (b)find the coordinate of a point Q that lies at the straight line such the |(OQ) ⃗ |= 3. I can't get this one, someone please guide me, thanks.
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  2. #2
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    Re: Vector

    Quote Originally Posted by alexander9408 View Post
    A straight line is given as 2(x+4) = -2(y-5) = -3(z-6),
    (a)determine the vector equation of the straight line. The answer I found is r = -4i +5j +6k + t(3i -3j +2k), is this correct?
    (b)find the coordinate of a point Q that lies at the straight line such the |(OQ) ⃗ |= 3. I can't get this one, someone please guide me, thanks.
    I get <-4,5,6>+t<3,-3,-2>. So check that 2k.
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  3. #3
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    Re: Vector

    Quote Originally Posted by alexander9408 View Post
    A straight line is given as 2(x+4) = -2(y-5) = -3(z-6),

    (a)determine the vector equation of the straight line. The answer I found is r = -4i +5j +6k + t(3i -3j +2k), is this correct?
    With t= 1, r= -4i+ 5j+ 6k+ 3i- 3j+ 2k= -i+ 2j+ 8k or the point (-1, 2, 8). 2(x+ 4)= 2(-1+ 4)= 6, -2(y- 5)= -2(2- 5)= 6, and -3(z- 6)= -3(8- 6)= -6. Those are NOT equal so the point (-1, 2, 8) on your line is not on the given line. They are not the same line.

    (b)find the coordinate of a point Q that lies at the straight line such the |(OQ) ⃗ |= 3. I can't get this one, someone please guide me, thanks.
    I presume you know that the distance from Q= (x, y, z) to O= (0, 0, 0) is \sqrt{x^2+ y^2+ z} so, setting that equal to 3 and squaring, x^2+ y^2+ z^2= 9.

    Since all of 2(x+ 4), -2(y+ 5), and -3(z- 6) are equal, set them all equal to t: 2(x+ 4)= t so x= -4+ (1/2)t. -2(y+ 5)= t so y= -5- (1/2)t. -3(z- 6)= t so z= 6- (1/3)t. Replace x, y, and z in x^2+ y^2+z^2= 9 to get a single equation in t.
    Last edited by HallsofIvy; October 6th 2012 at 01:27 PM.
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  4. #4
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    Re: Vector

    Hello, alexander9408!

    \text{A straight line is given as: }\:2(x+4) \:=\: \text{-}2(y-5) \:=\: \text{-}3(z-6)

    \text{(a) Determine the vector equation of the straight line.}

    To get the equations in Symmetric Form, divide by -6:

    . . \frac{x+4}{3} \;=\;\frac{y-5}{-3} \;=\;\frac{z-6}{-2}

    We have a point (\text{-}4,5,6) and the direction vector \vec v \:=\:\langle 3,\text{-}3,\text{-}2\rangle

    Write the vector equation.




    \text{(b) Find the coordinates of a point }Q\text{ on the line such that: }\,|\overrightarrow{OQ}| = 3.

    There are two solutions.

    I used the Parametric Form of the line: . \begin{Bmatrix}x &=& \text{-}4 + 3t \\ y &=& 5 - 3t \\ z &=& 6 - 2t \end{Bmatrix}

    Then: . |\overrightarrow{OQ}| \,=\,3 \quad\Rightarrow\quad \sqrt{(\text{-}4+3t)^2 + (5-3t)^2 + (6-2t)^2} \;=\;3

    . . 16-24t + 9t^2 + 25 - 30t + 9t^2 + 36 - 24t + 4r^2 \;=\;9

    . . 22t^2 - 78t + 68 \;=\;0 \quad\Rightarrow\quad 11t^2 - 39t + 34 \;=\;0

    . . (11t-17)(t-2) \:=\:0 \quad\Rightarrow\quad t \;=\;\tfrac{17}{11},\,2


    If t \,=\,\tfrac{17}{11}:\;Q(x,y,z) \;=\;\left(\tfrac{7}{11},\,\tfrac{4}{11},\,\tfrac{  32}{11}\right)

    If t\,=\,2:\;Q(x,y,z) \;=\;(2,\,\text{-}1,\,2)
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